Question

How does one verify `tanx+cosx/(1+sinx)=secx`?

`tanx+cosx/(1+sinx)=secx`

Answer 1

See below

Explanation:

Using:
`tanx=sinx/cosx`
`sin^2x+cos^2x=1`
`1/cosx= secx`

Start:

`tanx+cosx/(1+sinx)= secx`

`sinx/cosx+cosx/(1+sinx)= secx`

`(sinx(1+sinx))/(cosx(1+sinx))+(cosx*cosx)/((1+sinx)cosx)= secx`

`(sinx+sin^2x)/(cosx+cosxsinx)+(cos^2x)/(cosx+cosxsinx)= secx`

`(sinx+sin^2x+cos^2x)/(cosx+cosxsinx)= secx`

`(sinx+1)/(cosx+cosxsinx)= secx`

`cancel((sinx+1))/(cosxcancel((1+sinx)))= secx`

`1/cosx= secx`

`secx= secx`

Answer 2

1.) tan x = `((sin x)/"cos x")`
2.) plug it in: `((sin x)/"cos x")` + `cos x/ (1+sin x)`

3.) both sides must have a common denominator so:
`((sin x)(1+sin x))/((cos x)(1+sin x))` + `((cos x)(cos x))/((cos x)(1+sin x))`

4.) foil the equation (distribute):
`(sin x + sin^2 x)/(cos x +cosxsin x)` + `(cos^2 x)/(cos x + cosxsin x)`

5.) add numerators:
`(sin x + sin^2 x + cos^2 x)/(cos x + cosxsin x)`

6.) then: `cos^2 x + sin^2 x = 1`

`(1+sin x)/(cos x(1+sin x)`

7.) cancel out `1+sinx`

8.) left with: `1/cos x`

9.) simplify: `sec x`

Answer 3

`tanx+(cosx)/(1+sinx)`

`(sinx)/(cosx)+(cosx)/(1+sinx)`

`((sinx)(1+sinx))/((cosx)(1+sinx))+((cosx)(cosx))/((1+sinx)(cosx))`

`((sinx+sin^2x))/((cosx)(1+sinx))+((cos^2x))/((1+sinx)(cosx))`

`(sinx+sin^2x+cos^2x)/((1+sinx)(cosx))`

Remember `sin^2x+cos^2x=1` that's the pythagorean identity.

`(sinx+1)/((1+sinx)(cosx))`

`cancel((1+sinx))/(cancel((1+sinx))(cosx))`

`1/cosx`

`secx`

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Explanation:

With any fancy trig transformation, we can start from the left or the right. As for me, I want to start with complicated left and turn it into the right.
I took a guess at the strategy to use. On the right, secx is a fraction (it's 1 over cosx). So I should make the left into a fraction too. Let's start by turning tanx into a fraction (tanx=sinx/cosx). And then combine the two terms into a single fraction. Hopefully that fraction should simplify out. In fact it does, if you remember your identities. And it eventually gets to secx.